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Chapter 4: Problem 2

Determine the values of \(c\) at which \(f^{\prime}\) changes from positive to negative, or from negative to positive. $$ f(x)=x^{3}-x^{2}-x+2 $$

Short Answer

Expert verified
\(c = -\frac{1}{3}\) and \(c = 1\).

Step by step solution

01

Find the derivative

First, we need to find the derivative of the function \(f(x)\). The function is given as \(f(x) = x^3 - x^2 - x + 2\). Apply the power rule to each term to get the derivative:\[ f'(x) = 3x^2 - 2x - 1 \]
02

Solve f'(x) = 0

Next, we find the critical points by solving \(f'(x) = 0\). Set the derivative equal to zero:\[ 3x^2 - 2x - 1 = 0 \]This is a quadratic equation. Use the quadratic formula to solve for \(x\):\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]where \(a = 3\), \(b = -2\), and \(c = -1\). Substitute these values into the formula:\[ x = \frac{{2 \pm \sqrt{{(-2)^2 - 4 \times 3 \times (-1)}}}}{6} \]\[ x = \frac{{2 \pm \sqrt{{4 + 12}}}}{6} \]\[ x = \frac{{2 \pm \sqrt{16}}}{6} \]\[ x = \frac{{2 \pm 4}}{6} \]Thus, the solutions are \(x = 1\) and \(x = -\frac{1}{3}\).
03

Test intervals around critical points

Now that we have the critical points \(x = 1\) and \(x = -\frac{1}{3}\), we need to determine where \(f'(x)\) changes signs. Choose test points in the intervals defined by these critical points: 1. Interval \((-\infty, -\frac{1}{3})\), test with \(x = -1\)2. Interval \((-\frac{1}{3}, 1)\), test with \(x = 0\)3. Interval \((1, \infty)\), test with \(x = 2\)Calculate \(f'(x)\) for each test point:- For \(x = -1\), \(f'(-1) = 3(-1)^2 - 2(-1) - 1 = 3 + 2 - 1 = 4\) (positive)- For \(x = 0\), \(f'(0) = 3(0)^2 - 2(0) - 1 = -1\) (negative)- For \(x = 2\), \(f'(2) = 3(2)^2 - 2(2) - 1 = 12 - 4 - 1 = 7\) (positive)
04

Determine sign changes

Based on the test interval evaluations:- \(f'(x)\) changes from positive to negative at \(x = -\frac{1}{3}\).- \(f'(x)\) changes from negative to positive at \(x = 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivatives
In calculus, the derivative of a function represents the rate at which the function's value changes as its input changes. It's like finding the slope of a curve at any given point. For the function \( f(x) = x^3 - x^2 - x + 2 \), the derivative is calculated using the power rule, which involves differentiating each term separately and summing up the results.
To find the derivative, \( f'(x) \), apply:
  • For \( x^3 \), derivative is \( 3x^2 \).
  • For \( -x^2 \), derivative is \( -2x \).
  • For \( -x \), derivative is \( -1 \).
  • The derivative of a constant, such as 2, is zero.
Therefore, the derivative is \( f'(x) = 3x^2 - 2x - 1 \). Understanding derivatives is essential because they tell us how the function behaves, like where it increases or decreases.
Critical Points
Critical points occur where the derivative of a function is zero or undefined. These points are crucial because they can indicate where a function has a local maximum or minimum.
For the function \( f(x) \), we found the derivative \( f'(x) = 3x^2 - 2x - 1 \) in the previous section. Setting \( f'(x) = 0 \) identifies the critical points, leading to the equation \( 3x^2 - 2x - 1 = 0 \).
Solving this quadratic equation gives us the critical points \( x = 1 \) and \( x = -\frac{1}{3} \), which are potential points where the function's behavior might change, signifying peaks, troughs, or points of inflection.
Sign Changes
Sign changes in the derivative help locate intervals where the original function \( f(x) \) is increasing or decreasing. By selecting test points within intervals defined by critical points, we observe how \( f'(x) \) behaves.
Consider the intervals based on critical points:
  • Interval \( (-\infty, -\frac{1}{3}) \) with test point \( x = -1 \): \( f'(-1) = 4 \) (positive).
  • Interval \( (-\frac{1}{3}, 1) \) with test point \( x = 0 \): \( f'(0) = -1 \) (negative).
  • Interval \( (1, \infty) \) with test point \( x = 2 \): \( f'(2) = 7 \) (positive).
The sign change from positive to negative at \( x = -\frac{1}{3} \) suggests a local maximum, while the change from negative to positive at \( x = 1 \) suggests a local minimum. Understanding where these changes occur helps in sketching the function's graph and analyzing its behavior.
Quadratic Formula
The quadratic formula is a boon in calculus when tackling equations of the form \( ax^2 + bx + c = 0 \). It's a systematic method for finding solutions or the "roots" of quadratic equations.
For the equation \( 3x^2 - 2x - 1 = 0 \), we identify:
  • \( a = 3 \)
  • \( b = -2 \)
  • \( c = -1 \)
Substitute these into the quadratic formula:\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]This leads to computing the solutions:
  • Calculate the discriminant: \( (-2)^2 - 4 \times 3 \times (-1) = 16 \).
  • Find \( x = \frac{{2 \pm 4}}{6} \) resulting in \( x = 1 \) and \( x = -\frac{1}{3} \).
These roots are precisely the critical points where the sign of the derivative could change, critical for understanding the function鈥檚 graph.

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